Manifold #6. Topological Manifolds매니폴드 #6. 위상 매니폴드

These are my notes on XylyXylyX’s What is a Manifold? lectures. This post covers lesson 6 of the playlist, Topological Manifolds. Using homeomorphisms, compactness, and connectedness from the previous post, we now define a topological manifold. It is the first step toward the final goal, differentiable manifolds.

Conditions on the space

A topological manifold starts from a topological space $(X, \mathcal{T}_X)$. The space must first have three properties.

The first is Hausdorff: any two distinct points can be separated by open sets that do not overlap. The second is second countable: there is a basis made of countably many open sets, so every open set is a union of elements of that basis. The third is paracompact, a form of compactness: every open cover can be refined into a finer open cover so that every point has a neighborhood meeting only finitely many elements of the finer cover. The lecture says the details of paracompactness are not the key point, and encourages us to read about it once we understand compactness.

All three are topological properties, and together they make the space metrizable. That means we can give the space a distance, a rule that assigns a number at least 0 to each pair of points. We want to build a model of the real world where we measure distances, so we need this. Hausdorff distinguishes points clearly, and second countability and paracompactness keep the open sets under control. Most topological spaces in the world lack these properties, so not every space can be a topological manifold. (Paracompactness follows automatically from the locally Euclidean property below together with Hausdorff and second countable, which is why some definitions of a manifold drop the third condition. This explanation was added by me and was not covered in the lecture.)

Locally Euclidean

The other thing we need is Euclidean space $\R^n$. It is the set of $n$-tuples of real numbers, given the usual topology that has open balls as a basis. Now we have two topological spaces, the space $X$ under study and $\R^n$.

The first question that comes to mind is whether $X$ is homeomorphic to $\R^n$. If it is, we can do everything in $\R^n$ and come back, so the work ends there. Such spaces are interesting and are topological manifolds, but we want something weaker and more interesting. The whole space need not be homeomorphic to Euclidean space, as long as each point looks like $\R^n$ nearby. This is called locally Euclidean.

Let us see it concretely. Take a point $P$ of $X$ and an open set $U_P$ containing $P$, called an open neighborhood. If we collect only the elements of $\mathcal{T}_X$ that lie inside $U_P$, their unions and intersections stay inside $U_P$, the whole is $U_P$, and we can add the empty set. So $U_P$ itself becomes a topological space. This is called the subspace topology, inherited from $X$. Open sets of $\R^n$ work the same way. Since all these open sets are topological spaces, we can now talk about homeomorphisms between them.

We ask whether there is a homeomorphism $f_P$ from $U_P$ onto an open set of $\R^n$. This is a question about one neighborhood, not all of $X$, and for a homeomorphism we only need to confirm that such a function exists. If it works at $P$, we ask whether it works for the neighborhood $U_Q$ of another point $Q$, and we label it $f_Q$ to tell them apart. If we can find such a homeomorphism around every point, these neighborhoods form an open cover of $X$. The neighborhoods may overlap, and in fact they must, because open sets need overlaps to cover every point.

One thing to be careful about. $f_P$ and $f_Q$ go to different copies of $\R^n$. They cannot be laid on top of each other in one plane, and each goes to its own. The $n$-tuple where $f_P$ puts the point $P$ is the coordinates of $P$. Coordinates exist because $\R^n$ is defined as a set of ordered tuples. If we can cover every point with such neighborhoods and homeomorphisms, $X$ is locally Euclidean, and together with Hausdorff, second countable, and paracompact it is a topological manifold.

Q. Do $f_P$ and $f_Q$ land in overlapping regions of the same plane?
A. Each lands in a separate copy of $\R^n$. A point in both neighborhoods has a different coordinate in each of the two planes, and the relation between the two coordinates comes in the next post.

Example: the sphere

The most famous example is the sphere. Take the surface of a ball floating in space, and give it the topology whose basis is the places where open balls of $\R^3$ meet the surface. A ball that misses the surface gives the empty set, and a ball that contains the whole sphere gives the entire surface. This is the topology inherited from $\R^3$.

The sphere is compact. However we build an open cover, it has a finite subcover. The plane $\R^2$ is not compact. Tile the whole plane with slightly overlapping open rectangles and we need infinitely many, and no finite number covers the plane. Since compactness is a topological property, there is no homeomorphism between the sphere and the plane. No function carries the whole sphere onto the plane.

But if we remove one point from the sphere, say the south pole, the story changes. Around the south pole we can build open sets that shrink toward it, covering everything except the pole. This cover has no finite subcover, so the sphere without the south pole is not compact. In fact, the sphere without that point is homeomorphic to the plane. Picture sticking a knife into the missing point and spreading the sphere out over the whole infinite plane.

Other open parts of the sphere can also be sent to open sets of $\R^2$. The whole sphere is open too, but it cannot be sent, because then the whole sphere would be homeomorphic to the plane. The lecture points out that the whole sphere being both open and closed may be what sets it apart from the other open sets. (Concretely, it goes like this. If the whole sphere were homeomorphic to an open set $W$ of the plane, $W$ would be compact and open. The plane is connected, so its only nonempty subset that is both open and closed is the whole plane, but the plane is not compact, a contradiction. This explanation was added by me and was not covered in the lecture.) This is the signature of a topological manifold. The whole sphere need not be homeomorphic to a plane, as long as every point has a neighborhood homeomorphic to an open set of the plane.

Building charts by stereographic projection

Let us build this correspondence by hand. Looking at the sphere from the side, we see a circle, and the equator appears as a horizontal line. The plane $\R^2$ is this line, and we are looking at a cross-section of what happens in three dimensions.

Put the north pole at $(0, 1)$ and the south pole at $(0, -1)$. First take the open set $U_P$ that omits the south pole (in the lecture’s notation, $X$ minus the south pole $Q$). Draw the line from the south pole through a point $(x, y)$ of the circle, and let the image of the point be where this line meets the equator line. This construction is called stereographic projection. The north pole goes exactly to the center, points on the equator stay where they are, points of the northern hemisphere go inside the circle, and points of the southern hemisphere go outside. The closer we get to the south pole, the farther out the image goes without bound, and the south pole itself has no line, so it is undefined. The point we shoot from is what that chart cannot cover.

In formulas, the line from the south pole through $(x, y)$ is $(tx, -1 + t(1+y))$, and its $y$ coordinate is 0 when $t = 1/(1+y)$, so the coordinate on the equator line is

Shooting from the north pole in the same way gives $v = x/(1-y)$. For the sphere in three dimensions the same formula reads $(x, y, z) \mapsto (x/(1+z),\; y/(1+z))$. Conversely, from a point of the plane draw a line to the south pole and find where it meets the sphere. There is exactly one such place, so the inverse exists. The map is a bijection, onto, and continuous in both directions, so it is a homeomorphism. The lecture says the computation is easy with trigonometry or algebra, and the formulas for this construction were added by me.

Next we do the same thing again with the north pole removed. This gives an open set omitting the north pole and one more homeomorphism, shot from the north pole. These two open sets cover the whole sphere. A pair of a neighborhood and a homeomorphism is called a chart, and a collection of charts that covers the whole sphere is called an atlas. A chart is a picture of some region, just as maps bound into a book make an atlas. The sphere is a good example because it needs two charts.

The toy below lets you do the same on the circle, the cross-section of the sphere. Drag a point on the circle to see the coordinate $u$ of chart 1, shot from the south pole, and the coordinate $v$ of chart 2, shot from the north pole.

Q. Can one chart cover the whole sphere?
A. The whole sphere is compact and the plane is not, so the whole sphere cannot be homeomorphic to a single open set of the plane. That is why we remove one pole at a time and cover with two charts.

Q. Is there a relation between the two coordinates $u$ and $v$?
A. For a point on the circle $x^2 + y^2 = 1$, so $uv = x^2 / (1-y^2) = 1$. At every point except the two poles, one coordinate is the reciprocal of the other. (This computation was added by me and was not covered in the lecture. The lecture only previews that this relation is handled by transition functions in the next lesson.)

A counterexample: a line with a branch

The lecture also gave a classic counterexample that cannot be a manifold. Take one open line without endpoints, with another line branching off from a point in its middle. Put it inside $\R^2$ and give it the topology inherited from $\R^2$. Call this shape $X$.

Try sending it to $\R^1$. Parts far from the branching point go to intervals with no trouble. The trouble is near the branching point. If we send the branching point alone to a single point, then a single point is not an open set in $\R^1$, and continuity breaks. If we try $\R^2$ instead, a piece of line is not an open set in $\R^2$, so that does not fit. The same goes for other dimensions. This shape cannot be a topological manifold. The lecture notes this much and skips the detailed proof. (There is a clearer way to see it. Remove the branching point from a small enough neighborhood of it, and three pieces remain. Removing a point from an interval leaves two, and removing a point from a small ball of $\R^n$ with $n \ge 2$ leaves one. The number of pieces is a topological property, so no $\R^n$ is homeomorphic to it. This explanation was added by me and was not covered in the lecture.)

Summary

Gathering this post, a topological manifold is a Hausdorff, second countable, paracompact topological space in which every point has a neighborhood homeomorphic to an open set of $\R^n$. The sphere as a whole is not homeomorphic to a plane but is locally so, which makes it a topological manifold. A pair of a neighborhood and a homeomorphism is a chart, and a collection of charts covering the space is an atlas.

The next post starts where two charts overlap. On the sphere, the two chart regions overlap everywhere except the two poles. The function between the coordinates a point has in the two planes is a transition function, and it is the key to defining a differentiable manifold.